3.616 \(\int \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=226 \[ \frac{\left (3 a^2-8 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{4 \sqrt{b} d}+\frac{(-b+i a)^{3/2} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{3 a \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{(b+i a)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d} \]

[Out]

((I*a - b)^(3/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + ((3*a^2 - 8*b^2)*Arc
Tanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(4*Sqrt[b]*d) + ((I*a + b)^(3/2)*ArcTanh[(Sqrt[I*
a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + (3*a*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/(4
*d) + (Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2))/(2*d)

________________________________________________________________________________________

Rubi [A]  time = 1.49657, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3570, 3647, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ \frac{\left (3 a^2-8 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{4 \sqrt{b} d}+\frac{(-b+i a)^{3/2} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{3 a \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{(b+i a)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(3/2),x]

[Out]

((I*a - b)^(3/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + ((3*a^2 - 8*b^2)*Arc
Tanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(4*Sqrt[b]*d) + ((I*a + b)^(3/2)*ArcTanh[(Sqrt[I*
a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + (3*a*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/(4
*d) + (Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2))/(2*d)

Rule 3570

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n)/(f*(m + n - 1)), x] + Dist[1/(m + n - 1), Int[(a +
b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a^2*c*(m + n - 1) - b*(b*c*(m - 1) + a*d*n) + (2*a*b
*c + a^2*d - b^2*d)*(m + n - 1)*Tan[e + f*x] + b*(b*c*n + a*d*(2*m + n - 2))*Tan[e + f*x]^2, x], x], x] /; Fre
eQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && GtQ[n
, 0] && IntegerQ[2*n]

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d
*Tan[e + f*x])^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3655

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x
]}, Dist[ff/f, Subst[Int[((a + b*ff*x)^m*(c + d*ff*x)^n*(A + B*ff*x + C*ff^2*x^2))/(1 + ff^2*x^2), x], x, Tan[
e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} \, dx &=\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{1}{2} \int \frac{\sqrt{a+b \tan (c+d x)} \left (-\frac{a}{2}-2 b \tan (c+d x)+\frac{3}{2} a \tan ^2(c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{3 a \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{1}{2} \int \frac{-\frac{5 a^2}{4}-4 a b \tan (c+d x)+\frac{1}{4} \left (3 a^2-8 b^2\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\\ &=\frac{3 a \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{5 a^2}{4}-4 a b x+\frac{1}{4} \left (3 a^2-8 b^2\right ) x^2}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{3 a \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{\operatorname{Subst}\left (\int \left (\frac{3 a^2-8 b^2}{4 \sqrt{x} \sqrt{a+b x}}-\frac{2 \left (a^2-b^2+2 a b x\right )}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{3 a \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac{\operatorname{Subst}\left (\int \frac{a^2-b^2+2 a b x}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (3 a^2-8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=\frac{3 a \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac{\operatorname{Subst}\left (\int \left (\frac{-2 a b+i \left (a^2-b^2\right )}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{2 a b+i \left (a^2-b^2\right )}{2 \sqrt{x} (i+x) \sqrt{a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (3 a^2-8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 d}\\ &=\frac{3 a \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac{\left (i (a-i b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (i+x) \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac{\left (i (a+i b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left (3 a^2-8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{4 d}\\ &=\frac{\left (3 a^2-8 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{4 \sqrt{b} d}+\frac{3 a \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac{\left (i (a-i b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{i-(-a+i b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{\left (i (a+i b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{i-(a+i b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}\\ &=\frac{(i a-b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{\left (3 a^2-8 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{4 \sqrt{b} d}+\frac{(i a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{3 a \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\\ \end{align*}

Mathematica [B]  time = 6.10458, size = 854, normalized size = 3.78 \[ \frac{2 a \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)} \left (\frac{b \tan (c+d x)}{a}+1\right )^2 \left (\frac{3 \sqrt{a} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{8 \sqrt{b} \sqrt{\tan (c+d x)} \left (\frac{b \tan (c+d x)}{a}+1\right )^{5/2}}+\frac{1}{4} \left (\frac{1}{\frac{b \tan (c+d x)}{a}+1}+\frac{3}{2 \left (\frac{b \tan (c+d x)}{a}+1\right )^2}\right )\right )-i \left ((a+i b) \left (-\frac{2 \sqrt{a} \sqrt{b} \sqrt{\frac{b \tan (c+d x)}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a+b \tan (c+d x)}}-\frac{2 \sqrt [4]{-1} (a+i b) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{-a-i b}}\right )-2 b \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)} \left (\frac{b \tan (c+d x)}{a}+1\right ) \left (\frac{\sqrt{a} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{2 \sqrt{b} \sqrt{\tan (c+d x)} \left (\frac{b \tan (c+d x)}{a}+1\right )^{3/2}}+\frac{1}{2 \left (\frac{b \tan (c+d x)}{a}+1\right )}\right )\right )}{2 d}+\frac{2 a \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)} \left (\frac{b \tan (c+d x)}{a}+1\right )^2 \left (\frac{3 \sqrt{a} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{8 \sqrt{b} \sqrt{\tan (c+d x)} \left (\frac{b \tan (c+d x)}{a}+1\right )^{5/2}}+\frac{1}{4} \left (\frac{1}{\frac{b \tan (c+d x)}{a}+1}+\frac{3}{2 \left (\frac{b \tan (c+d x)}{a}+1\right )^2}\right )\right )-i \left (2 b \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)} \left (\frac{b \tan (c+d x)}{a}+1\right ) \left (\frac{\sqrt{a} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{2 \sqrt{b} \sqrt{\tan (c+d x)} \left (\frac{b \tan (c+d x)}{a}+1\right )^{3/2}}+\frac{1}{2 \left (\frac{b \tan (c+d x)}{a}+1\right )}\right )-(i b-a) \left (\frac{2 \sqrt{a} \sqrt{b} \sqrt{\frac{b \tan (c+d x)}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a+b \tan (c+d x)}}+\frac{2 \sqrt [4]{-1} (i b-a) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{a-i b}}\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(3/2),x]

[Out]

(2*a*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]*(1 + (b*Tan[c + d*x])/a)^2*((3*Sqrt[a]*ArcSinh[(Sqrt[b]*Sqrt[
Tan[c + d*x]])/Sqrt[a]])/(8*Sqrt[b]*Sqrt[Tan[c + d*x]]*(1 + (b*Tan[c + d*x])/a)^(5/2)) + (3/(2*(1 + (b*Tan[c +
d*x])/a)^2) + (1 + (b*Tan[c + d*x])/a)^(-1))/4) - I*(-2*b*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]*(1 + (b
*Tan[c + d*x])/a)*((Sqrt[a]*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(2*Sqrt[b]*Sqrt[Tan[c + d*x]]*(1 +
(b*Tan[c + d*x])/a)^(3/2)) + 1/(2*(1 + (b*Tan[c + d*x])/a))) + (a + I*b)*((-2*(-1)^(1/4)*(a + I*b)*ArcTanh[((-
1)^(1/4)*Sqrt[-a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a - I*b] - (2*Sqrt[a]*Sqrt[b]*Arc
Sinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/Sqrt[a + b*Tan[c + d*x]])))/(2*d) + (
2*a*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]*(1 + (b*Tan[c + d*x])/a)^2*((3*Sqrt[a]*ArcSinh[(Sqrt[b]*Sqrt[T
an[c + d*x]])/Sqrt[a]])/(8*Sqrt[b]*Sqrt[Tan[c + d*x]]*(1 + (b*Tan[c + d*x])/a)^(5/2)) + (3/(2*(1 + (b*Tan[c +
d*x])/a)^2) + (1 + (b*Tan[c + d*x])/a)^(-1))/4) - I*(2*b*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]*(1 + (b*T
an[c + d*x])/a)*((Sqrt[a]*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(2*Sqrt[b]*Sqrt[Tan[c + d*x]]*(1 + (b
*Tan[c + d*x])/a)^(3/2)) + 1/(2*(1 + (b*Tan[c + d*x])/a))) - (-a + I*b)*((2*(-1)^(1/4)*(-a + I*b)*ArcTanh[((-1
)^(1/4)*Sqrt[a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[a - I*b] + (2*Sqrt[a]*Sqrt[b]*ArcSin
h[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/Sqrt[a + b*Tan[c + d*x]])))/(2*d)

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Maple [B]  time = 0.353, size = 1345912, normalized size = 5955.4 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(3/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \tan \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^(3/2)*tan(d*x + c)^(3/2), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(3/2)*(a+b*tan(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Exception raised: RuntimeError